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### History and Applications of the Natural Logarithm

```
Date: 03/02/98 at 11:45:35
From: Paula Fornwalt
Subject: history of e

We have been using e in continuous growth problems, where we are
examining the growth potential of natural disease populations. I'm so
surprised at how often this number comes up in other applications,
though. Where did e come from? Who first derived it? Why is it so
common in the field of biology?
```

```
Date: 03/02/98 at 12:32:09
From: Doctor Rob
Subject: Re: history of e

For its history, see the following URL:

Earliest Uses of Symbols for Constants, by Jeff Miller
http://jeff560.tripod.com/constants.html

John Napier, the inventor of logarithms, is credited with discovering
this constant. Leonhard Euler is credited with popularizing the use of
the letter e for this number.

There are two reasons for its frequent appearance in mathematical
contexts.

The first is that it is the limit, as n grows without bound, of

(1 + 1/n)^n.

This particular definition lends itself to problems involving
continuous compounding of interest. Think of adding 1/n of the current
total to that current total n times.

The second is that e is the only constant such that the slope of the
graph of y = e^x at 0 is 1; or, in other words, the function e^x is
the solution of the differential equation dy/dx = y, with initial
conditions y = 1 when x = 0. Similarly, the solution of dy/dx = a*y,
with initial conditions y = 1 when x = 0, is e^(a*x).

This is the biological connection. You are talking about the
instantaneous rate of growth of something being proportional to the
amount of the something. That something can be a population of
bacteria in a culture, or a population of cockroaches in a garbage
dump. Here, "a" represents the fraction of them reproducing at any
given time x. If "a" is negative, it could represent the fraction of
organisms dying at any given time.

This is the model for radioactive decay of radium, uranium, and so on.
This is also the approximate model for inflation -- exponential
growth.

The way to calculate e is by using as many terms as necessary for the
accuracy you need in the following infinite series:

e = 1 + 1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*5) + 1/(2*3*4*5*6)
+ ...

The terms shown above are already enough to obtain a value for e that
is roughly 2.718.

-Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Definitions
High School History/Biography
High School Interest
High School Logs
High School Transcendental Numbers

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